What is the value of 4a cos3θ(d^2y/dx^2) if x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)?(a) [1 + (d^2y/dx^2)^2]^3/2(b) [1 – (d^2y/dx^2)^2]^3/2(c) [1 + (d^2y/dx^2)^2]^1/2(d) [1 – (d^2y/dx^2)^2]^1/2I got this question during an interview.My doubt is from Second Order Derivative topic in portion Limits and Derivatives of Mathematics – Class 11

What is the value of 4a cos3θ(d^2y/dx^2) if x = a sin2θ(1 + cos2θ)

1.	What is the value of 4a cos3θ(d^2y/dx^2) if x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ)?(a) [1 + (d^2y/dx^2)^2]^3/2(b) [1 – (d^2y/dx^2)^2]^3/2(c) [1 + (d^2y/dx^2)^2]^1/2(d) [1 – (d^2y/dx^2)^2]^1/2I got this question during an interview.My doubt is from Second Order Derivative topic in portion Limits and Derivatives of Mathematics – Class 11
Answer» The correct OPTION is (a) [1 + (d^2y/dx^2)^2]^3/2 For explanation I would say: Here, x = a sin2θ(1 + cos2θ) and y = a cos2θ(1 – cos2θ) => x = 2A cos^2θsin2θ and y = 2a sin^2θcos2θ Now DIFFERENTIATING x and y with respect to θ we GET, dx/dθ = 2a[cos^2θ2cos2θ + sin2θ2cos2θ cos2θ] = 4a cosθ (cosθ cos2θ – sinθ sin2θ) = 4a cosθ cos(θ + 2θ) = 4a cosθ cos3θ dy/dθ = 2a[cos2θ2cosθ sinθ + sin^2θ (-2sin2θ)] = 4a sinθ (cosθ cos2θ – sinθ sin2θ ) = 4a sinθ cos(θ + 2θ) = 4a sinθ cos3θ Thus, dy/dx = (dy/dθ)/(dx/dθ) = (= 4a cosθ cos3θ)/( 4a sinθ cos3θ) = tanθ So, (d^2y/dx^2) = d/dx(tanθ) = sec^2θdθ/dx = sec^2θ(1/(dx/dθ)) = sec^2θ1/(4a cosθ cos3θ) Or, 4a cos3θ (d^2y/dx^2) = sec^3θ = (sec^2θ)^3/2 = (1 + tan2θ)^3/2 As, dy/dx = tanθ So, 4a cos3θ(d^2y/dx^2) = [1 + (d^2y/dx^2)^2]^3/2

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